Answer
The pH of this buffer is equal to $3.40$.
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 4 \times 10^{- 4})$
$pKa = 3.398$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{1}{1}$
- 1: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{1}{4 \times 10^{-4}} = 2500$
- $ \frac{1}{4 \times 10^{-4}} = 2500$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.398 + log(\frac{1}{1})$
$pH = 3.398 + log(1)$
$pH = 3.398 + 0$
$pH = 3.398 \approx 3.40$